Question

find the nature of roots the following quaric equation big the real roots exist fineld them 2x+6x +3=o​

Answer 1

2x2 - 6x + 3 = 0

Comparing this equation with ax2 + bx + c = 0, we get

a = 2, b = -6, c = 3

Discriminant = b2 - 4ac

= (-6)2 - 4 (2) (3)

= 36 - 24 = 12

As b2 - 4ac > 0,

Therefore, distinct real roots exist for this equation

Refer to the attachment for the roots....